Question

10 g of ice at $0^{\circ }C$ absorbed 5460 J of heat to melt and change into water at ${50}^{\circ }C$. Calculate the specific latent heat of fusion of ice. Given specific heat capacity of water is 4200 ${kg}^{-1}{K}^{-1}$

Answer 1

Mass of ice = 10g = 0.01kg
Amount of heat energy absorbed,Q= 5460J
We have to find the specific latent heat of fusion of ice.
Specific heat capacity of water = $4200Jk{g}^{-1}{K}^{-1}$
Amount of heat energy required by 10g (0.01kg) of water at $0^{o}C$ to raise its temperature by ${50}^{o}C=0.01\times 4200\times 50=2100J$.

Let Specific latent heat of fusion of ice =$LJ{g}^{-1}$.
Then,
Q = mL + mc∆T5460
J =10 × L + 2100J
$L=336J{g}^{-1}$.